Fundamental Theorem of Algebra:
Introduction and Warm Up Algebra
Bren Calculus Workshop
Nathaniel Grimes
Bren School of Environmental Science & Management
Last updated: Sep 16, 2024
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Workshop Objectives
Shake off the math/schoolwork dust
Equip students with the math skills to succeed in all Bren Courses
Learn how valuable math is to environmental science
Build collaborative environment, crucial for success at Bren
We will be using team based learning
Science and policy are done collaboratively
TBL shows better learning outcomes than traditional lecturing
We’ll be doing a TBL Lite*, because I know orientation week is busy and school hasn’t officially started
Team Formation
Live coding demonstration
Team Expectations
Support and encourage each other
Communicate between all group members
Learn by teaching
Complete in class team assessments
Feel free to work with anyone on out of class exercises
Describe to your team what is the purpose of environmental science?
How would you go about solving environmental problems?
Math is an important tool in Environmental Science
Math to solve
Math is science’s foundation for finding evidence
Math verifies positive statements
Math to Communicate
Math is a language
Policy people need to know how to support their arguments
Math helps identify weakness in models, methods, and interpretation of results
If you don’t understand it, how can you make a decision on it?
Math at Bren
In classes:
ESM 201 Lokta-Volterra Models \[ \begin{align} \frac{dN_1}{dt}&=r_1N_1\left(\frac{K_1-N_1-\alpha N_2}{K_1}\right)\\ \frac{dN_2}{dt}&=r_2N_2\left(\frac{K_2-N_2-\beta N_1}{K_2}\right) \end{align} \]
ESM 222 Groundwater transport of absorbed contaminant \[ \frac{\partial C}{\partial t}=\left(\frac{D}{R}\frac{\partial^2t}{\partial x^2}\right)-\left(\frac{v}{R}\frac{\partial C}{\partial x}\right)-\frac{k}{R}C \]
In Research
We might be underestimating social cost of carbon because of adaptation (Dr. Jeon Bren PhD)
My own research in impacts from fishery insurance programs
Rules of Algebra
Never change an equation, we rewrite into more useful forms
Manipulate BOTH sides of an equation with the SAME THING
Order of Operations (aka PEDMAS)
(P)aranthesis (E)xponents (D)ivide (M)ultiply (A)dd (S)ubtract [PEDMAS]
PEDMAS important for what order to manipulate equations
\[ 4*(y-4)+(x+1)^2=z \]
If I give you x and y, how would you solve this equation?
Often times we want flexible equations
Prices are important in economics, but not always available for environmental goods.
How do we get prices if we know quantity?
\[ \require{cancel} \begin{aligned} Q&=\frac{(400-P)}{80} &\text{Isolate P in terms of Q} \\ \end{aligned} \]
Often times we want flexible equations
Prices are important in economics, but not always available for environmental goods.
How do we get prices if we know quantity?
\[ \require{cancel} \begin{aligned} Q&=\frac{(400-P)}{80} &\text{Isolate P in terms of Q} \\ 80Q&=\frac{(400-P)\cancel{80}}{\cancel{80}} &\text{ Multiply both sides by 80} \\ 80Q-400&=\cancel{400}-\cancel{400} -P &\text{ Subtract both sides by 400} \\ -1(80Q-400)&=-P(-1) &\text{Multiply both sides by -1} \\ 400-80Q&=P &\text{Flip terms for simplicity} \end{aligned} \]
It’s easy to make mistakes while doing algebra. Practice makes perfect
Solve all in terms of \(x\)
\[ 3x+2=10x-12 \]
\[ 4-3(2x+1)=8-\frac{3x}{2} \]
\[ 3(x+7a)-5=b+2(c-4x) \]
Practice Solutions
\[
\small
\begin{aligned}
3x+2&=10x-12 \\
3x+2+12&=10x-12(+12) \\
3x-3x+14&=10x-3x \\
14&=7x \\
x&=2
\end{aligned}
\]
\[ \small \begin{aligned} 4-3(2x+1)&=8-\frac{3x}{2}\\ 4-3-6x&=8-\frac{3x}{2} \\ 1-6x&=8-\frac{3x}{2} \\ 2-12x&=16-3x \\ -9x&=14 \\ x&=\frac{-14}{9}\\ \end{aligned} \]
\[ \small \begin{aligned} 3(x+7a)-5&=b+2(c-4x)\\ 3x+21a-5&=b+2c-8x \\ 11x+21a-5&=b+2c \\ 11x&=5+b+2c-21a\\ x&=\frac{5+b+2c-21a}{11} \end{aligned} \]
Exponents make algebra WAY tougher
\(x^n=x*x*x*x... (\text{n-times})\)
Many environmental variables follow exponential formulas like decay and growth
You might find yourself trying to solve equations like \(x^{\frac{3}{4}}=3x^{\frac{5}{3}}\)
Use the same principles of algebra and the properties of the right to manipulate
Rules and Properties of Exponents
\[ \begin{aligned} x^0&=1 & x\ne0 \\ x^{-n}&=\frac{1}{x^n} &x\ne0 \\ x^ax^b&=x^{a+b} \\ \frac{x^a}{x^b}&=x^{a-b}\\ (x^a)^b&=x^{ab}\\ (\frac{x}{y})^&a=\frac{x^a}{y^a}\\ (xy)^a&=x^ay^a\\ (\sqrt[n]{x}=a)&\to x=a^n \\ x^\frac{1}{n}&=\sqrt[n]{x}\\ x^\frac{m}{n}&=(x^{\frac{1}{n}})^m=(\sqrt[n]{x})^m \end{aligned} \]
Polynomials describe more complex relationships through exponents
\[ \large \begin{align} a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \end{align} \]
For example, polynomials often represent real world data better than a linear function
\[ \begin{aligned} CV&=b_0+b1\text{ row}+b_2 \text{ column}+ b_3\text{ row}^2\\ &+b_4\text{ column}^2+b_5\text{ row x column} +b_6\text{ elevation} \end{aligned} \]
The Fundamental Theorem of Algebra helps us solve for unknowns
\[ \large \begin{align} a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \end{align} \]
Fundamental Theorem of Algebra:
Every nth degree polynomial has exactly n zeros (solutions)
You were already finding zeroes!
\[ 3x+2=10x-12 \]
Finding nth degree polynomial zeros are much harder
Focus on 2nd degree polynomials
What do I mean by roots or solutions?
What happens if \(x=4\) or \(x=-3\) in \((x-4)(x+3)\)?
Try the same thing with \(x^2-x-12\)
The only way to get \(x^2-x-12\)=0 is for x to either be 4 or -3
4 and -3 are said to be the roots of the equation
(F)irst (O)utside (I)nside (L)ast
Second degree polynomials can be written as a multiplication of their “roots” aka solutions
We can expand using FOIL!
Multiply each term in the adjacent polynomial and add together
\[ \begin{align*} \text{First}&=x*x & \text{Outside}&=3x & \text{Inside}&=-4x & \text{Last}&=-12 \\ x^2-&x-12=0 & &\text{Add all terms together} \end{align*} \]
Quadratic Formula solves 2nd degree polynomials
For any second degree polynomial
\[ ax^2+bx+c=0 \]
The solution to can be found using the quadratic formula
\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]
\[ 4x^2+x-14=0 \]
\[ 256-\sqrt{44}x^2+.23x=10 \]
What happens if \(b^2-4ac\) is negative in the quadratic formula
Expand \((3x-6)(2x+1)\)
Solutions
\[ 4x^2+x-14=0 \]
\[ \begin{aligned} a=4 \\ b=1 \\ c=-14 \end{aligned} \]
Solutions
\[ 256-\sqrt{44} x^2+.23x=10 \]
\[ \begin{aligned} a=\sqrt{44} \\ b=.23 \\ c=246 \end{aligned} \]
This will probably be a nasty calculation, but that is what computers are for. The order does not matter, only that the a corresponds to the square term, the b to the 1st degree term, and the c to the constant
Solutions
There are no real solutions, but imaginary solutions. Can still be useful (saddle path solutions, but that’s for another class)
Solutions
\[ \begin{align} (3x-6)(2x+1) \\ 6x^2+3x-12x-6 \\ 6x^2-9x-6 \end{align} \]
Graphs bring visual connection to math and data
Which looks better and is easier to understand?
How to use graphs
Graphs move in a rectangular coordinate system with two dimensions (axes)
x - axis (horizontal)
y- axis (vertical)
Axis units must be defined
We use point pairs to place data
(X,Y)
Where would (2,-1) go on the graph?
Graphs series of points to make lines
Most 2D functions can be shown on graphs
Two key ingredients to graphs
X-intercept
y-intercept
What are the intercepts of the polynomial function in red?
Slope-Intercept Form
Easiest model to describe linear relationship between two variables
\[ \huge \begin{align} \underbrace{y}_{\text{y-variable}}=\overbrace{m}^{\text{Slope}}\underbrace{x}_{\text{x-variable}}+\overbrace{b}^{\text{y intercept}} \end{align} \]
Slope-Intercept Form
Vertical Change per unit of Horizontal Change
\[ m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1} \]
Horizontal Slope: m=0
Vertical Slope: m is undefined
Rise over Run
Slope represents rate of change
Slope can be average or instantaneous
You’re team measured the concentrations of pesticides in a lake exposed to agricultural runoff. The intern in charge of finishing the calculations ran off for the weekend leaving you all to finish their work. They left behind the following equation describing the total amount of pesticides in the lake if runoff is stopped from the farm by a new policy incentive reducing pesticides use:
\[ \begin{aligned} &y=(8-2t)(t+2) \\ &\text{Where } y \text{ is pesticde concentration in ppb} \\ &\text{and } t \text{ is time in years} \end{aligned} \]
Work with your team to discuss conceptually how you would solve the following tasks.
Write out the intern’s work in a more useful equation
How long will it take for the pesticide concentration in the lake to reach zero? Since the equation is a polynomial describe why one solution is more applicable than the other.
Present your findings (Choose between a graph or table)
What is the average change in concentration from year 0 to year 4?
Explain to your client why concentrations might behave the way they were modeled.
Let’s FOIL out the equation so it becomes easier to graph.
\[ \begin{aligned} y&=(-2t+8)(t+2) \\ y&=\overbrace{-2t^2}^{\text{First}}-\overbrace{4t}^{\text{Outside}}+\overbrace{8t}^{\text{Inside}}+\overbrace{16}^{\text{Last}}\\ y&=-2t^2+4t+16 \end{aligned} \]
Use the Quadratic Formula
\[ \begin{aligned} 0&=-2t^2+4t+16 \\ 0&=\frac{-4\pm\sqrt{4^2-4(-2)(16)}}{2(-2)}\\ 0&=\frac{-4\pm\sqrt{16+128}}{-4} \\ &t=4\text{, } t=-2 \end{aligned} \]
Use the factors
\[ \begin{aligned} 2t-8=0 \\ t=\frac{8}{2}\\ t=4 \end{aligned} \]
t=seq(-2,4)
y=-2*t^2+4*t+16
df<-data.frame(t=t,y=y)
p<-df %>%
ggplot(aes(x=t,y=y))+
geom_line(color="#003660",linewidth=3)+
labs(x="Years",y="Pesticide Concentration")+
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0),limits = c(-2,22))+
theme_classic()+
theme(text = element_text(size = 28))
pp2<-df %>%
filter(t>=0) %>%
ggplot(aes(x=t,y=y))+
geom_line(color="#003660",linewidth=3)+
labs(x="Years",y="Pesticide Concentration")+
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0),limits=c(-2,22))+
geom_hline(yintercept = 0)+
annotate("segment",x=0,xend=4,y=16,yend=0,color="#FEBC11",linewidth=2)+
theme_classic()+
theme(text = element_text(size = 28))
p2Use rise over run:
\[ \frac{\Delta y}{\Delta x}=\frac{0-16}{4-0}=-4 \]
Pesticides are removed from the lake at an average rate of 4 ppb per year
Pesticide concentrations might initially increase in the lake from residual particles in the soil being washed into the lake. Then a mixture microbial activity and other chemical process reduce the pesticides to more inert components (You will learn the actual answer in ESM 202)